Answer
• ∠B = 50º
,
• AB = 7.832
,
• CB = 5.035
Step-by-step explanation
The triangle ABC is shown below:
As it is a right triangle, we can use trigonometric functions to solve it. To know the hypotenuse (AB), we can use the cosine function:
![\cos(x)=\frac{adjacent\text{ side}}{hypotenuse}](https://img.qammunity.org/2023/formulas/mathematics/college/ongnrppkonhgqwzzm39azdc3b6oamwqbvv.png)
In our case, the adjacent side is b and the hypotenuse is AB. Then, by replacing our expressions we get:
![\cos(A)=(b)/(AB)](https://img.qammunity.org/2023/formulas/mathematics/college/338hule8ul446enjd3v7pgr8u51c0q5uke.png)
Next, by replacing the values and solving for AB we get:
![\cos(40\degree)=(6)/(AB)](https://img.qammunity.org/2023/formulas/mathematics/college/a67g8jslnimvijxdo9mxkcr5lyijxwply0.png)
![AB=(6)/(\cos(40\degree))](https://img.qammunity.org/2023/formulas/mathematics/college/wzha38nbisdkm0tmzpg5ryw7yak5e0my6d.png)
![AB\approx7.832](https://img.qammunity.org/2023/formulas/mathematics/college/sh4eeqaf2w6fazlkghvgfikizhspaztd3g.png)
As we have two sides, we can use the Pythagorean Theorem to find side CB:
![AB^2=b^2+CB^2](https://img.qammunity.org/2023/formulas/mathematics/college/yc73v28mgakm3nunqmrkop2p3xfmr5gvdg.png)
![AB^2=b^2+CB^2](https://img.qammunity.org/2023/formulas/mathematics/college/yc73v28mgakm3nunqmrkop2p3xfmr5gvdg.png)
Next, we can solve for CB (the side that we are lacking) as follows:
![CB=√(AB^2-b^2)](https://img.qammunity.org/2023/formulas/mathematics/college/qzftqcxmqi64a72lkou3yfmsfqkcrolosk.png)
![CB=√(7.832^2-6^2)](https://img.qammunity.org/2023/formulas/mathematics/college/ln8wkdqxzihn54x81wp09sb8xddlk81pim.png)
![CB=√(7.832^2-6^2)\approx5.035](https://img.qammunity.org/2023/formulas/mathematics/college/hu386uy2czchz5kyp1edtkosq99b9emx87.png)
Finally, as the addition of the interior angles of a triangle adds up to 180º, we can find ∠B as follows:
![\angle A+\angle B+\angle C=180\degree](https://img.qammunity.org/2023/formulas/mathematics/college/6hv1d76v6p5s7vrub2hbgr3ts6t7dpyxzh.png)
![40\degree+\angle B+90\degree=180\degree](https://img.qammunity.org/2023/formulas/mathematics/college/lcyc1kmrl6yb048lu6mcwi3g107q35sik8.png)
![\angle B=180\degree-90\degree-40\degree](https://img.qammunity.org/2023/formulas/mathematics/college/1vejywvmbx4nicnlk3hj3969ev66dvr392.png)
![\angle B=50\degree](https://img.qammunity.org/2023/formulas/mathematics/college/ms10093eqhqvochd05c6idenha0uu79246.png)